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Navigation of electric fish. Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electric field in the water around them. (See Figure $19.67 . )$ This field creates a potential difference of a few volts between the head and tail, which in turn
causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivities different from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to this current and can detect changes in it. The changes in the current allow the fish to navigate. (In the next chapter, we shall investigate how the fish might detect this current.) Since the electric field is weak far from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawater through which that field passes as a conducting tube of area 1.0 $\mathrm{cm}^{2}$ and having a potential difference of 3.0 $\mathrm{V}$ across its ends. The length of a Nile fish is about $20 \mathrm{cm},$ and the resistivity of seawater is 0.13$\Omega \cdot \mathrm{m} .$ (a) How large is the current through the tube of seawater? (b) Suppose the fish swims next to an object that is 10 $\mathrm{cm}$ long and 1.0 $\mathrm{cm}^{2}$ in cross-sectional area and has half the resistivity of seawater. This object replaces the seawater for half the length of the tube. What is the current through the tube now? How large is the change in the current that the fish must detect? (Hint: How are this object and the remaining water in the tube connected, in series or in parallel?)

So I am. Exchange chromatography is executed by making the reversible exchange of irons between irons present and solution on iron exchange resin. So we tend to use this to separate similar irons. And so in the first part where it's calculating the total amount of solution fed into the column 36 7.2 liters. So in part B, the expression of a volumetric flow rate for solution a be a T think was not point for liters per minute. The VA 33.6 minutes multiplied by t was able to not point not 11 90 so we can apply the mass balance in solution A M put a is equal to accumulation at a and we are able thio. The salmon that has Time T varies from not to 33.6 minutes and V A varies from zero to V a bracket. L The volume of Solution A becomes 6.72 liters on. We can calculate the amount of sodium chloride there's an A. C. L is equal to 392.72 grams and a c l. And so we can give an expression for environment for a lower rate of solution be that is tu minus no point no. 11 19 that is equal to be be dark t. So the total volume of solution a 60.48 liters total volume solution by solution be not at all number of moles of tres fad. Three points. There are two full mall on some possible potentials for error. So pumped be might have failed, causing a flow of one mole of sodium chloride to the column during the illusion period. Programs method chosen to execute in the chromatography could be incorrect. Perhaps the column was packed with the wrong reason, maybe arising. That has reduced affinity compared to the correct resin.

For this question. We're looking at the nuclear pore complex specifically at the poor structure. We're being told we have F G repeats and but in vitro at a concentration at least 50 million more, please form a gel. Our first part is to ask if it's reasonable to expect them to form a gel in vivo as well. And for this we're given some data. So first thing you want to do is write down the information you've been given, we've been told, is 5000 repeats and that they are found in poor but is 35 9 m in diameter and 30 nanometers any length. So how are we going to work out the concentration of these repeats in this this volume? Well, we're going to work out the belonging we're working with first, and it says it's cylindrical. So we're going to use the formula for the volume of a cylinder, which is pi r squared pH. So the volume of the poor is going to be pi times R squared. So at 17.5 squared times 30 and this is going through this, our volume of 28,849 remember Nanometers cubes. We have to keep track of our unit, sir. So you've got Nanometers Cube now, right? So this is the volume, and we have 5000 repeats. What does that really mean? Well, we're going to work out how many malls 5000 repeats is, So we're gonna have 5000, and we're gonna just divided by aggregator is constant. So six, um, times 10 to 23. And this gives us not, um, scripts. US 5/6 times tend to be minus 20. I'll leave in this form now, so make it easier to use in future equations. But you can get a rough idea of what we're looking at here. So we have five of six times sentiments. 20 months, right? Perfect. We have our amount. We have our volume. To get the density or we have to do is divide one by the other. So we're going to work with our There were 5000 over six times five of six times 10 to the minus 20 like so. And we're going to divide this by our area. So we're going to divide this by 28,849 nanometers cubes and note that we're still using nanometers cubed. We have to deal with that. If we wanted to go from from nanometers to, say, centimeters, that would be a factor of seven sense for seven but its cubes There's gonna be a factor of 10 to 21. And furthermore, we are also, uh, we're also working with moles. We're going to end up in mill malls, but anyway, onwards. So we're also going to multiply our end results by 10 to 21 to bring us out of nanometers and into centimeters cubed. So all of this gives us 289. And if we do not 0.289 and if we convert it into minimal, most polite by thousands, we get 289 Miller malls. So it's all so this is our overall concentration. And as you can see, it is more than sufficient to form a gel. We're looking for at least 50. We're getting 287 so we have we have plenty, right? That's part a. Now we're gonna be looking at part B. Alright. For part B, we're looking at the diffusion of um, a couple of molecules. First, we're looking at X M V p. A protein, and then we're looking at a tagged M V P. But also it has important attached foot. And if we look at the diagrams and book, we see that without important, it does not pass through a membrane and nuclear membrane. But with important, it does no. Is the fusion of this important bound tagged protein fast enough to account for the efficient flow of materials and were given some more data? Let's just write down the information we have. We have the diffusion ordinates D, which is not 0.1 micrometers cubed per second. We also have the equation for diffusion, which is helpful, and we want to see how long it takes to move through a poor, uh, that is 30 nanometers. So we have X equals 30 nanometers, Correct. So the equation is going to be when we rearrange it, we're going to get t equal to X squared over two D. So we have t is equal to 30 nanometers squared, 30 squared, and we'll keep in mind. We're looking at nanometer strip, um, so divided by D, which is not quite one micrometers square per second. And so we have nanometers micrometers here. We want to We want to change this. Um, So what I'm going to do is I'm going to more supply by Tend to be minus three on this hot here. There we go. So that this will now be in nanometers squared per second. So we have nanometers squared over nanometer square per second, which is going to give us what we need. And this gives us not point. Not, not for five seconds. So will change out to, uh, 4.5 milliseconds. So again, the most important bit for part B is to make sure you change your not 0.1, uh, micrometers in 29.1 times 10 to 9 to three nanometers to keep everything working with units. But this gives us 4.4 point five milliseconds, and if he wants to make a comment on that, that seems reasonable. That's very fast. Um, and so we would expect that to match in fever because in Viva, we will be looking at a rate of about five milliseconds to import a protein through a nuclear pore

There are certain steps to draw a graph. The first step is G c toe in so it but job second will be in. So it ext. Why Scotto graph Third will be input X y values that is concentration on. Would it then fourth will be goto design view on select FX type graph draw using the given data. The graph octane is not linear for the case, eh? It really represents like this for option B. If you draw a graph between log value on the concentration along with voltage, you'll get a linear So far, option A. It is normally new option B. It is linear for option C. We draw a graph off log concentration off corporate too positive bosses voltage as described in bart a. Sew the graft we obtained boards world age by log corporate too positive will be representing somewhere like this. The equation line will be represented as my equal 0.0 to nine x last 0.3 full in option E In the above equation for straight line by his bold age, while X is logged value off concentration, So subsidy of the given by value into the equation to determine the value off X for the both the cases, so the concentration of corporate O positive is zero point 053 on dhe 0.853 respectively.

Problem 98 party. This is the escape of cells and this is an old act as negative terminal And this is the catholic fact as positive terminal terminal. And this is a directional flow of electrons And this is also the directs no flow up electron in order surrogate where this is solvable. This is short of breath. Okay, this is the platinum metal. Very nearly good. Have he plus two and Happy Plus three. And in this liquid. Yes, sure, losing that is cat hard is at plus um and no food minus and mm close to movement of an iron in this direction where as moment of car tire in this direction. So this is final answer for party. No, in part of me in part to be we need this is the yeah escaped the process that occur at the atomic level at the surface of an order. This is the platinum follicular labor and this is the iron after that. After reacting. This is another another diagram with sign iron is good. 10 10 Occidental instead prompt plus two, two plus three. So this answer for part to be now in part C. Mhm. In part C. This is first we need to write half cell reaction in both standard and catalogued. So first we write habits and reactions occurring on cat or terminal. That is eman both food minus echoes plus eight x plus echoes plus five electron. It will give mm plus two a course plus poor edge to liquid and external reduction potential for this heart reaction is one point five months old now half sell the accent that occur on and or terminal. That is Iron Plus two. Air Coast gave iron plus three echoes Plus one electron. And each tender reduction potential for this reaction is close to zero point 771 World. No, we can. Right overall P. M. For this cell reaction Is in Articles two. He not reduction for cat hold minus in our production. For an order. Now simply put the values So we can write in articles two. One point I've one whole minus 0.771 Hole. That is equals two 0.74 old. This is final answer for part B. This is now in part C. We need to calculate E. M. F. Of herself when concentration of substances given. So in part today we can write AS B equals two. The note minus 0.0 592 divided by and into log concentration of F. E plus three to the power five into concentration of amen plus two, divided by concentration of um fee plus two with about five into concentration of a man or poor minus into Concentration of as plus to the Power eight. No value of and according to poison. You clearly see here Value of n equals 2 5 and P. It is given. Therefore Given Ph Equals two 0.0. That is mean that means Concentration of answer and equals to one. Now we simply put the value so we can write equals not value of in order. We already calculated in part C calculating Part C. That is zero 744-0.0592, divided by five log. Now put the values of concentration That is 2.5 Into 10. to the power -4 To the power five in group concentration of magnesium and that is 0.0 01 divided by concentration of Happy. Plus do that at zero 10 to the power pipe into concentration of one 50 into concentration. Or we want to the power it. Now. we can clearly see here value of Q. After calculation it's 6.5 point 510 into 10. to the power minus 17. No, we can write as equals to 0.74 minus 0.0592 by five, log 6.5- 10 to the Power -17. No listen, we can write 0.74 minus 0.0592, divided by five. Multiply. These values equals two minus 16 point 18 64 After calculation, we get 0.74. Hold bless 0.19 hole so EMF of cell equals two zero point 9 3. World. This is the final answer for part of the


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